It is clear that the form of \(\lambda \)  will depend on the transformation \(\phi _t\) .
In the standard pull-back case, it takes the form of the divergence of the velocity by which the domain moves, see the formula for \(\lambda _\Omega \)  at the bottom of page 21 of [1]}.
We now calculate \(\lambda \)  for the Piola transform, making use of the characterisation [2]}.
\(\frac{{\rm d}}{{\rm d}t} \int _{\Omega (t)} u(t)\cdot v(t)=&\frac{{\rm d}}{{\rm d}t}\int _{\Omega _0} (u\circ \Phi _t )\cdot (v \circ \Phi _t) \det (D\Phi _t)=\frac{{\rm d}}{{\rm d}t}\int _{\Omega _0} \left(D\Phi _t \phi _{-t} u\right) \cdot \left(D\Phi _t \phi _{-t} v\right)\\=&\int _{\Omega _0} D\Phi _t\frac{{\rm d}}{{\rm d}t} \left( \phi _{-t} u \right) \cdot \left( D\Phi _t \phi _{-t} v\right) + \left( D\Phi _t \phi _{-t} u\right)\cdot D\Phi _t\frac{{\rm d}}{{\rm d}t} \left( \phi _{-t} v \right)\\&+ \phi _{-t} u \cdot \frac{{\rm d}}{{\rm d}t}\left( D\Phi _t^T D\Phi _t \right) \phi _{-t} v.\nonumber \) 
